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Is it possible to supply the 25W Halogen bulb which is 4.16 Ampere and 6V with 4 in series combine with 3 sets of that in parallel D Cell Energizer/Duracell (Alkaline).

If yes, how long is the burning time?

Informations of Energizer D cell (Alkaline) is :
http://data.energizer.com/SearchResult.aspx

2007-06-11 05:04:47 · 2 answers · asked by senjaya_chang 1 in Science & Mathematics Engineering

2 answers

You need to be careful here. The nominal rating of a single alkaline D cell is 20.5 Ah, but this is at a current of only 25mA.

Your bulb is rated at 4.16A, so each branch of your parallel circuit will have to supply 4.16 /3 = 1.39A, and as this current passes through each of the cells in series, each cell also has to supply 1.39A.

Looking at the data sheet, these cells are not even rated for such a high current, and I would be surprised if you got more than an hour or so out of them before the light became unacceptably dim.

To make this work, you would need at least twice as many parallel branches (i.e. 24 cells) to reduce the current to 695mA. Looking at the graphs, I would guess that the cells should then give you around 10 hours' light.

In other words, Zn-MnO2 cells like these deliver a lot more energy if you take it out more slowly.

2007-06-11 05:55:38 · answer #1 · answered by rrabbit 4 · 0 0

3 sets of 4 D cells will produce 61.5 Amp-hours.

That's 15 hours of full light, and well over 20 hours of useful light -- maybe up to a full day of useful light (24 hours).

EDIT: after looking at the discharge curves in a little more detail, I'll downgrade my estimate to an upper limit of 15 hours of useful light with about 3-5 hours of full bright light.

.

2007-06-11 05:23:58 · answer #2 · answered by tlbs101 7 · 0 0

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